Two springs A and B fixed at the top and are stretched by $8 \mathrm{~cm}$ and $16 \mathrm{~cm}$…

Two springs A and B fixed at the top and are stretched by $8 \mathrm{~cm}$ and $16 \mathrm{~cm}$ respectively, when loads of $20 \mathrm{~N}$ and $10 \mathrm{~N}$ are suspended at the lower ends. The ratio of the spring constants of the springs $A$ and $B$ is
  1. $1: 1$
  2. $2: 1$
  3. $3: 1$
  4. $4: 1$

Solution

Spring constant, $K=\frac{F}{x}$ Given, $\mathrm{F}_{\mathrm{A}}=20 \mathrm{~N}, \mathrm{~F}_{\mathrm{B}}=10 \mathrm{~N}, \mathrm{x}_{\mathrm{A}}=8 \mathrm{~cm}$ and $\mathrm{x}_{\mathrm{B}}=16 \mathrm{~cm}$ $ \begin{aligned} & \therefore \frac{\mathrm{K}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{B}}}=\frac{\frac{\mathrm{F}_{\mathrm{A}}}{\mathrm{x}_{\mathrm{A}}}}{\frac{\mathrm{F}_{\mathrm{B}}}{\mathrm{x}_{\mathrm{B}}}}=\frac{\mathrm{F}_{\mathrm{A}}}{\mathrm{x}_{\mathrm{A}}} \times \frac{\mathrm{x}_{\mathrm{B}}}{\mathrm{F}_{\mathrm{B}}}=\frac{20}{8} \times \frac{16}{10}=4 \\ & \therefore \mathrm{K}_{\mathrm{A}}: \mathrm{K}_{\mathrm{B}}=4: 1 \end{aligned} $

Asked in: AP EAMCET 2023 (15 May Shift 1)

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